Let the 4-digit locker key be&nbsp;$abcd$, where&nbsp;$9 \ge a,b,c,d \gt 0$, and they are all unique.It is given that&nbsp;$a \le 3$ &amp; $\frac{b}{c}=a$It is also given that 'd' is the smallest number.Case (i): $a=1$Not possible, as&nbsp;'d' is the smallest number, 'a' cannot be 1.Case (ii): $a=2$Then 'd' can take only one value, i.e.&nbsp;$d=1$&nbsp;$b=2c$if $c=3$, then $b=6$if $c=4$, then $b=8$for $c\ge\ 5$, 'b' will not&nbsp;be a single-digit number.Hence, two cases are possible, i.e. (2631, 2841).Case (iii): $a=3$'d' can take two values, i.e.&nbsp;$d=1$ or $d=2$&amp; $b=3c$If d = 1, then 'c' can take only one value. i.e.&nbsp;$c=2$, then $b=6$If d = 2, then 'c' cannot take any value.Hence one case is possible, i.e. (3621)Amit has to try 3 different combinations. Option (E) is correct.

The first number Haruka entered is 1. From Rule-1, the series will be 2, 4, 8, 16 From Rule 2, 16 will be followed&nbsp;by 7 Like-wise, the series will be 1, 2, 4, 8, 16, 7, 14, 5, 10, 1, 2, 4,...... The series is repeating itself for every 9 iterations. Initially, input is 1 and after first iteration, result is 2.&nbsp; This implies, 9n = 1 9n + 1&nbsp;= 2 9n + 2&nbsp;= 4 . . . . 9n + 7&nbsp;= 5 9n + 8&nbsp;= 10 In the question, it is given that key is pressed 68 times.&nbsp; 68 = 9(7) + 5 68 is in the form of 9n + 5 This implies, 7 is the output. The answer is option A.

Option A) False, as it is mentioned no-one scored exactly 40 marks. Option B) False, as only 1 student&nbsp; or only 2 students can score above 40 and group average can be 40. &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Case i:- 44, 39, 39, 39 &amp;&nbsp;39. &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Case ii:- 42, 41, 39, 39 &amp; 39. Option C) False, &nbsp;student's scores&nbsp;could be&nbsp;44, 39, 39, 39 &amp; 39. Option D) False, student's scores could be 44, 43, 42, 32 &amp; 39. Option E) True, if scores of all students are more than 40. The average will be more than 40.

From the given information, we get that it is&nbsp;$8\ inch\times\ 8\ inch$&nbsp;square grid.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_3kIWiME.png">Total ways of selecting 2 squares out of 64 in&nbsp;$^{64}C_2$.Two squares with a common side can be selected in the following ways.(i) Horizontal Pairs.In the first row, R1, we can select 7 pairs of squares with a common side.They are&nbsp;(R1C1,R1C2),&nbsp;(R1C2,R1C3),.....(R1C7,R1C8).It applies to other rows as well.Hence the total number of squares =&nbsp;$7\times\ 8=56$(ii) Vertical Pairs.In the first column, C1, we can select 7 pairs of squares with a common side.They are (R1C1,R2C1), (R2C1,R3C1),.....(R7C1,R8C1).It applies to other columns as well.Hence the total number of squares = $7\times\ 8=56$The probability of two painted squares having a common side =&nbsp;$\frac{56+56}{^{64}C_2}$ = $\frac{112}{2016}$.Option (A) is correct.

Let 'm' be the Greatest common divisor and&nbsp;A, B, C, D, E, F &amp; G be the 7 unique numbers. When 'm' divides A, B, ......&amp; G, we get seven unique&nbsp;quotients. Let the quotients be a, b, c, d, e, f, and g. It is given, A + B + C + D + E + F + G = 1740 m(a + b + c + d + e + f + g) = 1740 $\ \ 1740=2^2\times3\times5\times29$ m(a + b + c + d + e + f + g) =&nbsp;$\ \ 1740=2^2\times3\times5\times29$ 29 is a prime number and cannot be factorised further. Therefore, largest possible value m can take is 60. Least possible sum of 7 unique numbers is 1 + 2 + 3 +...+ 7 = 28 Replacing 7 with 8, we will get 29. This implies 60(1 + 2 + 3 .....+ 6 + 8) = 1740 Largest possible G.C.D is 60. The answer is option B.

Let players at slips 1, 2 and 3 be a, b, and c, respectively.Let the speed of the ball be m.The trajectory of the ball is not specified in the question. But let's take it as straight line.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_xyOyH68.png">The distance between the 1st and 2nd slips is the same as the distance between the 2nd and 3rd slips.The length of 'ab' = The length of 'bc'. It implies Xb is the median of triangle Xac.It takes 3 sec, 4 sec and 5 sec for the ball to reach a, b and c, respectively.Hence, Xa = 3m, Xb = 4m and Xc = 5m.Using Apollonius's theorem,$2\times\ \left(\left(4m\right)^2+\left(ab\right)^2\right)=\left(3m\right)^2+\left(5m\right)^2$$16m^2+\left(ab\right)^2=17m^2$$ab=m$Hence,&nbsp;$ac=2m$Using the properties of triangle, the sum of two sides should be greater than the third side,But, Xa + ac = Xc3m + 2m = 5m.&nbsp;Hence, this arrangement is not possible.&nbsp;As this&nbsp;question has ambiguity, XAT officials awarded full marks to all candidates for this question.

$Let\ \sin^215^{\circ}\ =a\ and\ \sin^275^{\circ}\ =\ b$Then the given equation becomes,$\ \frac{\ a^3+b^3+6ab}{a^2+b^2+5ab}$&nbsp;=$\ \frac{\ \left(a+b\right)^3-3ab\left(a+b\right)+6ab}{\left(a+b\right)^2-2ab+5ab}$&nbsp;&nbsp;&nbsp;$\ \therefore\ \ a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$ ,&nbsp;$\ \therefore\ \ a^2+b^2=\left(a+b\right)^2-2ab$Using&nbsp;$\sin\left(\theta\right)\ =\cos\left(90-\theta\ \ \right)$we get,&nbsp;$\sin^275^{\circ\ }=\cos^215^{\circ }$now, a+b = $\sin^215^{\circ\ }+\sin^275^{\circ }$ = $\sin^215^{\circ\ }+\cos^215^{\circ\ }$ = 1.The equation becomes,&nbsp;$=\ \frac{\ 1-3ab+6ab}{1-2ab+5ab}\ =\ \frac{\ 1+3ab}{1+3ab}\ =1$

Let 'a' and 'b' be the speeds of Rahim and the speed boat in still water, respectively Let 'c' be&nbsp;the speed of the stream. Speed boat starts after 1 minute of Rahim's departure and crosses him after 4 minutes. It implies that the speed boat covered the same distance in 4 minutes that Rahim took 5 minutes. Hence, the ratio of upstream speeds of Rahim and Speed boat =&nbsp;$\frac{a-c}{b-c}=\frac{4}{5}$ Using Statement 1, we get $b=30\ kmph$. But, it's not sufficient to get the value of 'a'. Using Statement 2, we get&nbsp;$a-c=\frac{30}{3}=10\ kmph$. But it's also not sufficient&nbsp;to get the value of 'a'. using both statements, we get $\ \frac{\ a-c\ }{b-c}=\frac{4}{5}$ $\ \frac{\ 10\ }{30-c}=\frac{4}{5}$ $\ \ 50\ =120-4c$ $\ \ c\ =17.5$ $\ \ a=17.5+10=27.5\ kmph$

Let 'm' be the number of matches Guava played till now. They won '0.4m' matches.&nbsp;After&nbsp;playing another 'n' matches and wining&nbsp;all of them, their winning percentage will improve to 50.i.e,&nbsp;$\ \frac{\ 0.4m+n}{m+n}=0.5$&nbsp;&nbsp;&nbsp;$m=5n$Playing 15 more matches and wining all of them, their winning percentage will improve from 50 to 60.i.e $\ \frac{\ 0.4m+n+15}{m+n+15}=0.6$Solving, we get $6+0.4n\ =\ 0.2m$Substituting&nbsp;$m=5n$,we get&nbsp;$6+0.4n\ =\ n$&nbsp;$n=10$ &amp; $m =50$Hence, Guava club played 50 matches so far.&nbsp;The answer is option A.

Given that,&nbsp;$AD\bot\ BC$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$BE=3$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$BC=5$<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_qKF7w6x.png"> Using Pythagoras' theorem,&nbsp;&nbsp;&nbsp;$AD^2+BD^2=AB^2$ .....(1)&nbsp;&nbsp;&nbsp;$AD^2+DE^2=AE^2$......(2)&nbsp;&nbsp;&nbsp;$AD^2+DC^2=AC^2$......(3)(1) - (2) gives$BD^2+DE^2=AB^2-AE^2$$x^2-\left(3-x\right)^2=AB^2-AE^2$$AB^2-AE^2=6x-9$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$&nbsp;&nbsp;&nbsp;&nbsp;($CD=\left(5-x\right)$)$AB^2-AE^2+6CD=6x-9+30-6x$$AB^2-AE^2+6CD=21$

xat 2023 Complete Paper Solution | QADI

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